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Let $X$ be a topological space. Let $\Sigma$ be the set of irreducible components of $X$. Let $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$, $X_i,Y_j\in \Sigma $ for some index set $I,J$. $X_i$'s are distinct from each other. $Y_j$'s are distinct from each other.

I want an example such that $X$ has two distinct expression, i.e. there exist $\{X_i|i\in I\}\neq\{Y_j|j\in J\}$, such that $X=\cup_{i\in I} X_i=\cup_{j\in J} Y_j$.

Something we knew (but not helpful for this question): $X$ must not be one of the following case:

  1. $X$ is Noetherian, then it can be uniquely written as a union of finite distinct irreducible components.
  2. $X$ can be written as a union of finite distinct irreducible components, then all irreducible components of $X$ are in this expression, and expression is unique.
  3. $X$ is a scheme, since {irreducible components} 1:1 correspond {generic points}. Thus expression is unique.

Update: 4. As stated, if $X$ is Hausdorff, then every irreducible set is a single point. (Because $E$ irr.$\Leftrightarrow$ every two nonempty opens intersect.)

Thanks.

wxu
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  • I'm no expert on the topic, so I'm leaving this comment to be addressed to by greater minds; should this question be tagged under [algebraic-topology] as well/instead? – Asaf Karagila Jun 21 '11 at 15:45
  • I've added [algebraic-geometry], since this is the field of math in which irreducible components are important. – Jim Belk Jun 21 '11 at 15:58
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    It's worth mentioning that an example can't be Hausdorff. Any Hausdorff space with at least two points has the property that two points have disjoint neighborhoods, so those neighborhoods have closures which are not the entire space. So the only irreducible Hausdorff spaces are points and the irreducible decomposition of a Hausdorff space is trivially unique. – Qiaochu Yuan Jun 21 '11 at 16:09
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    More generally, an example can’t be a sober space. (I’m not familiar with schemes, but I suspect that I’m essentially just restating wxu’s (3).) – Brian M. Scott Dec 09 '11 at 18:37
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    @BrianM.Scott: yes, schemes are sober spaces. –  Dec 10 '11 at 22:20
  • Even more generally, an example can't be quasi-sober (every closed irreducible subset has at least one generic point). – Elías Guisado Villalgordo Aug 24 '23 at 18:18
  • More general point 2: an example cannot be a space with locally finitely many irreducible components (i.e., a space where each point has a neighborhood with finitely many irreducible components). – Elías Guisado Villalgordo Aug 24 '23 at 18:38
  • More general point 1: an example cannot be a locally Noetherian space (a space with an open cover of Noetherian topological spaces), for locally Noetherian implies locally with finitely many irreducible components. For instance, a classical prevariety over field $k$ (a locally ringed space over $\operatorname{Spec}k$ that is locally isomorphic to $V(I)\subset k^n$ with the sheaf of regular functions, where $I\subset k[x_1,\dots,x_n]$ is some ideal) is locally Noetherian. (If $k=\overline{k}$, then one's doing algebraic geometry; if $k$ is a real closed field, then one does real AlgGeom.) – Elías Guisado Villalgordo Aug 24 '23 at 18:39

2 Answers2

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This should work. Take the square $[0,1]\times [0,1]$ with this strange topology. A subset is open if it is open in the Zariski topology (no generic points!) of every vertical segment, and in the Zariski topology of the horizontal segment $[0,1]\times\{0\}$.

Then, it's clear that the irreducible components are the vertical segments and $[0,1]\times\{0\}$, but to cover the square it's enough to use the vertical ones.

Elías Guisado Villalgordo
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Giulio Bresciani
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  • I wrote here a proof of the claim that the irreducible components are the vertical segments and $[0,1]\times{0}$. It's just elementary general topology, although I don't see how to deduce the claim from any more general ideas. – Elías Guisado Villalgordo Aug 24 '23 at 17:27
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    Here is the way I think about this. Call special segments the vertical ones and the lower horizontal one. If C is closed irreducible and p is a point of C, either p is contained in a special segment contained in C or C\p is closed. Hence, either C is a point or it is a union of special segments. It is immediate to see that a closed, irreducible union of special segments is a special segment. – Giulio Bresciani Aug 27 '23 at 11:16
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$\def\spec{\operatorname{Spec}} \def\bbA{\mathbb{A}} \def\frp{\mathfrak{p}} \def\sV{\mathcal{V}}$Here's an example coming from algebraic geometry (inspired by G. Bresciani's example). Let $k$ be an infinite field.

Let $A=\dfrac{k[x,y_a\mid a\in k]}{((x-a)y_a\mid a\in k)}$ and $X=\spec A$.

We have surjections \begin{align*} A&\to\dfrac{A}{(y_a\mid a\in k)}\\ A&\to\dfrac{A}{(x-b,y_a\mid a\in k\setminus\{b\})},\quad b\in k, \end{align*} that give rise, respectively, to closed subschemes \begin{align*} \spec k[x]\cong H\to X\\ \spec k[y_b]\cong V_b\to X \end{align*}

Geometrically, $X$ is a copy of $\mathbb{A}_k^1\cong H$ to which we have glued the origin of $\mathbb{A}_k^1\cong V_a$ at each point $a\in H$. More precisely, on $k$-rational points, we have that $X(k)= k\times k$, where $H(k)$ and $V_a(k)$ correspond respectively to $k\times\{0\}$ and $\{a\}\times k$, as subsets of $X(k)$ (here, ‘$H$’ and ‘$V$’ stand for ‘horizontal’ and ‘vertical’; a drawing for $k=\mathbb{R}$ may elucidate why).

We note that $H$ plus the $V_a$'s cover $X$ (later, we will only need $X(k)=H(k)\cup\bigcup_{a\in k}V_a(k)$): suppose $z\in X$ does not lie in $H=\sV(y_a\mid a\in k)$. Denote $\frp\subset A$ to the associated prime of $z$. Then $y_b\notin\frp$, for some $b\in k$, whence $x-b\in\frp$ (since $(x-b)y_b=0$ in $A$). Thus, $x-a\notin\frp$ for all $a\in k\setminus\{b\}$; hence, $y_a\in\frp$ for all $a\not = b$. In other words, $\frp\supset(x-b,y_a\mid a\in k)$, i.e., $z\in V_b$.

We claim that the irreducible components of $X(k)$ are exactly $H(k)$ and $V_a(k)$, $a\in k$. Hence, we would have a counterexample. It suffices to verify the hypotheses of the lemma here (since $|k|=\infty$):

  1. $V_a(k)$ is closed and $V_a'(k):=D(y_a)(k)$ is open.

  2. The subspaces $H(k)$ and $V_a(k)$ have the cofinite topology (they are homeomorphic to $\bbA_k^1(k)$).

  3. We show that $H(k)$ is an irreducible component of $X(k)$: suppose $Z\subset X(k)$ is a closed irreducible subset containing $H(k)$. Note that $Z$ does not contain $V_a(k)$ (if it did, then $Z=V_a(k)\cup (Z\setminus V_a'(k))$ is a union of closed proper subsets); hence, $Z\cap V_a(k)$ is finite. We want to show that $Z=H(k)$. To look for a contradiction, suppose there is $z\in Z\setminus H(k)$, i.e., $z\in Z\cap V_a'(k)$, for some $a\in k$. It suffices to see that $Z\setminus\{z\}$ is closed in $X(k)$ (for $Z=(Z\setminus\{z\})\cup\{z\}$ and $X(k)$ is $\mathrm{T}_1$). Write $Z\cap V_a'(k)=\{(a,b_1),\dots,(a,b_n)\}$, and suppose $z=(a,b_1)$. Then $Z\setminus\{z\}=Z\cap \sV(y_a(y_a-b_2)\cdots(y_a-b_n))$ is closed.

¹Set-theoretically speaking, the intersection of $H$ with $V_a$ is $(a,0)$ (i.e., the prime $(x-a,y_a\mid a\in k)\subset A$), which is $k$-rational, so $H\cap V_a=(H\cap V_a)(k)=H(k)\cap V_a(k)$.

Elías Guisado Villalgordo
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